An Electrostatics Problem—C.E. Mungan, Fall 1999
The following wonderful little problem was fully discussed and analyzed on the PHYS-L email
list.
Serway’s 4th edition of Physics for Scientists and Engineers problem 57 of chapter 23 states in
part: Two small spheres of mass m are suspended from strings of length L that are connected at a
common point. One sphere has charge Q; the other has charge 2Q. Assume that the angles θ1 and
θ2 that the strings make with the vertical are small. How are θ1 and θ2 related?
Before reading further, try the problem yourself. It has a very simple solution. But then ask
yourself this: Why the restriction to small angles?
Simple solution: Invoke the condition of rotational equilibrium to conclude that the center of
mass of the system of two spheres must lie on a vertical line through the pivot point. Hence
θ1 = θ 2 . It is now trivial to solve for the common tension, angle, electrostatic force, and anything
else you like.
Another approach: The poser of the problem went on to ask whether it can be solved by invoking
the condition of translational equilibrium alone? The answer is yes and it nicely shows why we
require the restriction to small (in the sense of acute, it turns out) angles. Start with a general
diagram,
L
L
T1
θ1
θ3
θ2
θ3
T2
2Q
F
mg
Q
θ4
F
mg
where F is the electrostatic repulsive force between the two charges. The very hard way to do
this problem is to set up a coordinate system with x and y horizontal and vertical, and to take
components. Try it and see: Using the geometric fact that θ 4 = (θ2 − θ1 )/ 2 gives the messy
equation
⎛θ −θ
cot θ1 − cot θ2 = 2tan⎝ 2 1 ⎞⎠
(1)
2
after some elimination of unknowns among the four component equations. This is therefore a
good example for students who always want to orient coordinate systems this way. A much
better choice is to orient one axis of the coordinate system for each mass along the respective
string. The perpendicular force balance then gives
mg sin θ1 = Fsin θ 3
(2)
for the Q charge, and similarly
mg sin θ2 = F sin θ3
(3)
for the 2Q charge. Equating the left-hand sides of Eqs. (2) and (3) implies
sin θ1 = sin θ2 .
(4)
This has two solutions:
(i) θ1 = θ 2 as before; or
(ii) θ1 = 180˚−θ 2 , which implies that one sphere is directly across the circle from the other, thus
putting the center of mass at the pivot, as is acceptable albeit unstable to rotations of the system.
Of course this can only happen if F has a very large value, so that the string does not collapse.
This solution is however ruled out by the small-angle restriction, which thus does not necessarily
imply the small-angle approximation that you may have thought of from the problem wording.
Comments: Notice that the above two solutions are consistent with Eq. (1) and further that the
small-angle approximation to Eq. (1) does give solution (i). Nevertheless I am unable to prove
that (i) and (ii) are the general solutions to Eq. (1). Can you? I suspect that a graphical proof is
needed, as trig identities only gave me a huge mess.
It is also worth noting in hindsight that both solutions are in fact given by our initial simple
solution. After all, the symmetry of both the gravitational force and the Coulombic force merely
require that the horizontal distance from one mass to a vertical line through the pivot be equal
and opposite to that of the other mass. However, the more formal alternative approach spares one
from drawing hasty conclusions from symmetry, as one might easily be led to do. Shame, shame!